Question

A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0°, and 60.0°.

Answer 1

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

$vsin(\alpha) + gt = 0$

$t = \frac{-vsin(\alpha)}{g}$

So the vertical distance it travels within time t is

$y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}$

$y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}$

$y = \frac{-v^2sin^2(\alpha)}{2g}$

Similarly the horizontal distance it travels within time t is:

$x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}$

$x = \frac{-v^2sin(2\alpha)}{2g}$

We can pre-calcualte the quantity $\frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54$

So $y = 57.54sin^2(\alpha)$

$x = 57.54sin(2\alpha)$

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m